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A particle of mass m moves in a unidimensional potential field U= kx^(2)//2 (harmonic oscillator). Using the uncertainty principle, evaluate the minimum permitted energy of the particle in that field. |
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Answer» Solution :We write `P~DeltaP~(ħ)/(Deltax)~(ħ)/(x)` i.E., all FOUR QUANTITIES are of the same ORDER of magnirude. Then `E~~(ħ^(2))/(2 mx^(2))+(1)/(2)kx^(2)((ħ)/(x)-sqrt(mk)x)^(2)+ħsqrt((k)/(m))` Thus we GET an equilibrium situation (`E=` minimum) when `x=x_(0)=sqrt((ħ)/(sqrt(mk)))` and then `E=E_(0)~~ħsqrt((k)/(m))=ħ omega` Quantum mechanics gives `E_(0)=ħ omega//2` |
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